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I like to check my answers with wolframalpha, and this one's stubbornly coming up as false when set equal to its answer for the antiderivative, but I can't figure out where I'm going wrong.

Using the identity $\cos^2x = (1-\sin^2x)$, I rewrote the integral as:

$$\int \cos(x)(1-\sin^2x)^2(\sin^5x)\,dx$$

$$u = \sin x\\ du = \cos x\,dx\\ dx = du/\cos x$$

Canceling out the stray cosine, that turns into $\int(1-u^2)^2u^5 \, du$

Expanding, it's $\int (1-2u^2+u^4)u^5\,du = \int (u^5-2u^7+u^9)\,du$

$$= \frac{\sin^6 x}{6}-\frac{\sin^8 x}{4}+\frac{\sin^{10} x}{10}+C$$

Basically, does anybody see an error here? Is there an error?

vonbrand
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Calculistening helplessly
  • 625

2 Answers2

6

You could also rewrite in this way \begin{align} \cos^5(x)\sin^5(x)&=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^5 \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^5\\ &=\left(\frac{e^{2ix}-e^{-2ix}}{4i}\right)^5\\ &=\frac{1}{4^5i}\left(e^{10ix}-5e^{6ix}+10e^{2ix}-10e^{-2ix}+5e^{-6ix}-e^{-10ix}\right)\\ &=\frac{1}{2\cdot4^4}\left(\sin(10x)-5\sin(6x)+10\sin(2x)\right) \end{align}

Vincenzo Tibullo
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3

Another approach: $\frac{\sin 2x}2 =\sin x \cos x$ yielding:

$$\frac{1}{2^5}\int \sin^5 2x\,dx$$

Letting $u=2x$ then $du=2dx$ and we get:

$$\frac{1}{2^6}\int \sin^5 u\,du$$

Writing $\sin^5 u = (1-\cos^2 u)^2 \sin u$, and letting $v=\cos u$, you get:

$$-\frac{1}{2^6}\int (1-v^2)^2 \, dv$$

This yields:

$$-\frac{1}{2^6}\left(v-\frac{2}{3}v^3+\frac{1}{5}v^5+C\right)$$

Substitute back in $v=\cos 2x$.

Thomas Andrews
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